HAOI2011 problem a

与i分数相同的人数的区间[a[i] + 1, n - b[i]]
设f[i]表示确定了i个人后满足要求的最大人数
对于一个区间[l, r]，用map弄出它的个数num，f[r] = max(f[l - 1] + num)
其实不用排序，挂条链就可以了

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 10);

IL ll Read(){
    char c = '%'; ll x = 0, z = 1;
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    return x * z;
}

int n, f[_];
map < pair <int, int>, int > M;
vector <int> r[_];

int main(RG int argc, RG char *argv[]){
    n = Read();
    for(RG int i = 1; i <= n; i++){
        RG int a = Read(), b = Read();
        if(a + b + 1 > n) continue;
        if(++M[make_pair(a + 1, n - b)] == 1) r[n - b].push_back(a + 1);
    }
    for(RG int i = 1; i <= n; i++){
        f[i] = f[i - 1];
        for(RG int j = 0, l = r[i].size(); j < l; j++)
            f[i] = max(f[i], f[r[i][j] - 1] + min(i - r[i][j] + 1, M[make_pair(r[i][j], i)]));
    }
    printf("%d
", n - f[n]);
    return 0;
}